The Robinson-Foulds (RF) metric provides symmetric distance between two phylogenies as a sum of monophyletic groups present in one tree but not in the other. The normalised RF distance is given as $\frac{RF_i}{RF_{max}}$. Let's consider an example:

tr = rtree(10, rooted = FALSE)
tr2 = rtree(10, rooted = FALSE)
RF.dist(tr, tr2)
# [1] 12
RF.dist(tr, tr2, normalize = TRUE)
# [1] 0.8571429

The last row would mean that the trees differ by 86 % of their maximum distance. But how does one interpret the RF distance if it is weighted on branch lengths?

wRF.dist(tr, tr2)
# [1] 13.20986
wRF.dist(tr, tr2, normalize = TRUE)
# [1] 0.7600115

Is it 76 % of maximum tree distance given the sum of tree lengths of the compared trees?

  • $\begingroup$ You could look at the help from within R (try ?wRF or ?wRF.dist). If this doesn't give you an answer, the best way to be sure would be to look at the source code of this package. There may be ways to do it from within R: stackoverflow.com/questions/19226816/… $\endgroup$ – bli Oct 16 '17 at 9:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.